# Yet Another Delta-Epsilon

## Homework Statement

Prove that lim (1/x) x --> 0 does not exist, i.e., show that lim (1/x) x --> 0 = l is false for every number l.

0 < |x-a| < d
|f(x) - l| < E

## The Attempt at a Solution

The strange thing is, the first time through I got the same solution as Spivak, but looking over it again the logic seems downright wrong. Here's the solution, verbatim:

The function f(x) = 1/x cannot approach a limit at 0, since it becomes arbitrarily large near 0. In fact, no matter what d > 0 may be, there is some x satisfying 0 < |x| < d, but 1/x > |l| + E, namely, any x < min(d, 1/(|l| + E)). Any such x does not satisfy |(1/x) - l| < E.

Where is he getting the bold portion from?

I write |1/x - l| < E
|1/x| - |l| < |1/x - l| < E
|1/x| - |l| < E
|1/x| < E + |l|

How is he getting greater-than?

Office_Shredder
Staff Emeritus
Gold Member
I write |1/x - l| < E

Assuming this is true, then you're golden. But it's not going to be true for really small x, that's the point. So we see how he's doing a proof by contradiction: If the limit exists, you can't possibly get a greater than sign, but he gets a greater than sign. The way he gets it is by saying when x < min(d, 1/(|l| + E)), you get 1/x > |l| + E

vela
Staff Emeritus
Homework Helper
Where is he getting the bold portion from?
You're misreading what he wrote. He is merely asserting that there is an x such that 0<x<d and 1/x>|l|+E. Then he tells you how to find such an x.
I write |1/x - l| < E
|1/x| - |l| < |1/x - l| < E
|1/x| - |l| < E
|1/x| < E + |l|

How is he getting greater-than?
If the inequality you started with is true, then your conclusion is also true. If you go one step further and take the reciprocal of both sides, you get

$$|x| > \frac{1}{|l|+\epsilon}$$

Spivak is saying if you choose an x such that this last inequality is false, the one you started with will also be false.

So, he's essentially saying:

|1/x| < E + |l|

and then pointing out that, if d = 1/(|l| + E), this leads to a contradiction of that inequality, and therefore the whole thing violates the definition of a limit?

Oy! It's been almost a week and I'm still having trouble with these things.

Office_Shredder
Staff Emeritus
Gold Member
So, he's essentially saying:

|1/x| < E + |l|

He isn't saying that because it's not going to be true. But it's what you should keep in the back of your mind... IF l was the limit, you would get this inequality. But in fact we get an opposite inequality for small x. Therefore, l is not the limit. Since l was arbitrary, there is no limit

vela
Staff Emeritus
Homework Helper
No, he's stating if 0<|x|<min{δ, 1/(|L|+ε)}, then 0<|x|<δ and |1/x|>|L|+ε. The δ appears only because the definition of the limit requires that |x|<δ. So long as |x| is small enough, you can satisfy both conditions. The point is, regardless of what δ equals, you can always find an |x| between 0 and δ such that |1/x|>|L|+ε.

When you try to come up with a delta-epsilon proof, typically you work backwards. You assume the epsilon condition holds and find some condition on x from which you can choose an appropriate δ. In this case, you can't find such a δ because the inequality turns out backwards.

The strange thing is, the first time through I got the same solution as Spivak, but looking over it again the logic seems downright wrong.

I have those kinds of days too :) lol

Let me try again: because there are values for delta (e.g. <1/(|L| + E)) that violate the inequality that would result if the limit were true, then the limit is not true by contradiction?

vela
Staff Emeritus
Homework Helper
No, that's not right. Even when a limit exists, you can usually find a δ for which the ε-inequality won't hold: just use a big δ. The limit doesn't exist because there are no values of δ for which the ε-inequality holds.

No, that's not right. Even when a limit exists, you can usually find a δ for which the ε-inequality won't hold: just use a big δ. The limit doesn't exist because there are no values of δ for which the ε-inequality holds.

Yeah, I think I figured it out a few hours ago (I was doing another problem and realized what I said here was wrong). It's because we've found a d within which no x will work (all x results in an inconsistency, no matter how small; in fact, the smaller the x the worse the inconsistency becomes)?

vela
Staff Emeritus